MIT OpenCourseWare http://ocw.mit.edu 16.346 Astrodynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Lecture 18 Preliminary Orbit Determination Using Taylor Series #3.7 Taylor’s Series (1712) Brook Taylor (1685–1731) The vector r can be expressed as a Taylor series about r1 with time interval τ = t2 − t1 : τ2 τ3 r2 = r1 + τv1 + 2! r��1 + 3! r���1 + O(τ4) Approximate Solution of the BVP using Taylor’s Series Differentiate the series twice and use the equation of motion µr��2 + �2r2 = 0 where �2 = 3r2 Then r��2 = r��1 + τr���1 + τr���−�2r2 = −�1r11 Hence τ2 τ2 r2 = r1 + τv1 − 2! �1r1 + 3!(�1r1 − �2r2) Solve for τ v1 to obtain � 1 �1 �� 1 �2 � v1 = − τ r1 ++ τ r2 Problem 3–31 τ − 3 τ 6 valid to third order in the time interval τ = t2 − t1 . Another Method of Gibbs using Taylor Series Pages 136–137 r(t)= a0 + a1(t − τ )+ a2(t − τ )2 + a3(t − τ )3 + a4(t − τ )4 + O[(t − τ)5] where a0 , a1 , a2 ,...are the function and its derivatives evaluated at time t = τ . • Given r1 = r(t1), r2 = r(t2), r3 = r(t3) To determine p we have six equations in the six unknowns a0 , a1 , a2 , a3 , a4 , p:• r1 = a0 − a1τ3 + a2τ32 − a3τ33 + a4τ34 τ1 = t3 − t2 r2 = a0 where τ2 = t3 − t1 = τ1 + τ3 Also r3 = a0 + a1τ1 + a2τ12 + a3τ13 + a4τ14 τ3 = t2 − t1 �1(p − r1)=2a2 − 6a3τ3 +12a4τ32 µd2r�2(p − r2)=2a2 since where � == �(p − r) rdt2 �3(p − r3)=2a2 +6a3τ1 +12a4τ12 16.346 Astrodynamics Lecture 18 3������ ������ ����������� ����������� � ´ Etienne Bezout’s Theorem Appendix D Consider the system of linear algebraic system a1x + a2y + q1 =0 b1x + b2y + q2 =0 of two equations in two unknowns x and y . If a third equation is included, a1x + a2y + q1 =0 b1x + b2y + q2 =0 c1x + c2y + q3 =0 the system is now over-determined. A necessary and sufficient condition for the system to be consistent is that a1 a2 q1 b1 b2 q2 c1 c2 q3 =0 Since we require only the parameter p, we can choose p to make the system consistent 1 −τ3 τ32 −τ33 τ34 r1 1 0 0 0 0 r2 1 τ1 τ12 τ13 τ14 r3 0 0 2 −6τ3 12τ32 �1(p − r1) 0 0 2 0 0 �2(p − r2) 0 0 2 6τ1 12τ2 �3(p − r3)1 =0 Thus r1τ1(1 + �1A1) − r2τ2(1 − �2A2)+ r3τ3(1 + �3A3) p = τ1�1A1 + τ2�2A2 + τ3�3A3 where 12A1 = τ2τ3 − τ2 12A2 = τ1τ3 + τ2 12A3 = τ1τ2 − τ2 1 2 3 • To determine a we again have six equations in six unknowns with the last three equations created from the relation: d2 �11 dt2 (r 2)=2µr − a Problem 3–7 Then, in the same manner as before, we find 2 2 2µr1 r2 r3 a = − τ2τ3 (1 − 2�1A1)+ τ1τ3 (1 + 2�2A2) − τ1τ2 (1 − 2�3A3) 16.346 Astrodynamics Lecture 18Laplace’s Method (1780) • Given the unit vectors iρ1(t1), iρ2(t2), iρ3(t3) where r = ρ + d and ρ = ρ iρ • Determine ρ2 , dρ2/dt, d iρ2/dt from which we obtain r2 = ρ2 iρ2 + d2 dρd i2 ρ2 ddv2 = iρ2 + ρ2 + 2 dt dt dt From the second derivative of the vector ρ = ρ iρ d2 ρ d2iρ dρdid2ρ = ρ +2ρ + idt2 dt2 dt dt dt2 ρand the equations of motion d2 ρ d2r d2d µ µ µ µ = = d r = d (ρ + d) 2 2 2 3 3 3 3 ρD1 = µdρ 2 D = µ� 1 1 d3 − r� D2 (I)3 11 dt1 r � r3 − Dd3 3 (II) 2 = ρ2 + d2 +2� ρ(iρ · d) (III) 1. Solve (I) and (III) for r2 and ρ2 . dρ2. Use (II) to determine at time t2 . dt 16.346 Astrodynamics Lecture 18 dt dt−dt d−r d−rwe have �d2ρµ � dρdiρ d2iρ � 1 1 � dt2 + r3 ρ iρ +2dt dt + ρ dt2 + µr3 − d3 d = 0 Next, take the scalar product of this vector equation with the vector cross-product diρiρ × dt � d iρ d2iρ �� 1 1 �� d iρ � to obtain ρ iρ × dt · dt2 +µr3 − d3 iρ × dt · d =0 � �� � � �� � = D1 = D2 d2iρSimilarly, using iρ × dt2 dρ � d2iρ diρ �� 1 1 �� d2iρ � gives 2dt iρ × dt2 · dt +µr3 − d3 iρ × dt2 · d =0 � �� � ��� � = −D1 = D3 As a result, we have Laplace’s equations��� ��� ��� Lagrange’s Interpolation Formulas (1778) The vector iρ(t)can be expressed as a Taylor series iρ(t)= a0 +(t − t0)a1 + 1 (t − t0)2a2 + O(t − t0)3 2 Expanding about the time t2 gives iρ3 = a0 + τ1a1 + 12τ12a2 iρ2 = a0 iρ1 = a0 − τ3a1 + 21τ32a2 a0 = iρ a1 = diρ d2iρ a2 = dt2dt t=t0 t=t0 t=t0 Hence, we have τ1a1 + 1τ12a2 = iρ3 − iρ22τ3a1 − 21τ32a2 = iρ2 − iρ1 to be solved for a1 and a2 . The result is diρ dt d2iρ dt2 ���� t2����� t2 = − τ1 iρ1 + τ1 − τ3 iρ2 + τ3 iρ3τ2τ3 τ1τ3 τ1τ2 2 2 2 = iρ1 − iρ2 + iρ3τ2τ3 τ1τ3 τ1τ2 which are valid to second order in the time intervals where τ1 = t3 − t2 τ2 = t3 − t1 = τ1 + τ3 τ3 = t2 − t1 More accurate values for these derivatives can be obtained if more than three sets of observational data are available. The determination of the derivatives of the observational data is the greatest weakness in Laplace’s method of orbit determination. In fact, it is necessary to use additional observations to obtain any reasonable accuracy. 16.346 Astrodynamics Lecture